Question

the one - to - one functions g and h are defined as follows.
$g = {(-4, 8), (-1, -8), (4, -3), (8, 3)}$
$h(x)=\frac{x + 9}{5}$
find the following.
$g^{-1}(8)=square$
$h^{-1}(x)=square$
$(hcirc h^{-1})(2)=square$

Explanation:

Part 1: Find \( g^{-1}(8) \)

Step1: Recall inverse of a function

For a one - to - one function \( g \) defined as a set of ordered pairs \( (x,y) \), the inverse function \( g^{-1} \) maps \( y \) back to \( x \), i.e., if \( (x,y)\in g \), then \( (y,x)\in g^{-1} \).
We are given \( g=\{(-4,8),(-1, - 8),(4,-3),(8,3)\} \). We want to find \( g^{-1}(8) \), which is the \( x \)-value such that \( g(x) = 8 \). Looking at the ordered pairs in \( g \), when \( y = 8 \), \( x=-4 \).

Step1: Let \( y = h(x) \)

We have \( h(x)=\frac{x + 9}{5} \), so let \( y=\frac{x + 9}{5} \).

Step2: Solve for \( x \) in terms of \( y \)

Multiply both sides of the equation \( y=\frac{x + 9}{5} \) by 5: \( 5y=x + 9 \).
Then subtract 9 from both sides: \( x = 5y-9 \).

Step3: Replace \( y \) with \( x \) to get the inverse function

Since \( h^{-1}(x) \) is the inverse of \( h(x) \), we replace \( y \) with \( x \) in the equation \( x = 5y-9 \). So \( h^{-1}(x)=5x - 9 \).

Step1: Recall the property of inverse functions

For a function \( h \) and its inverse \( h^{-1} \), the composition \( h\circ h^{-1}(x)=x \) for all \( x \) in the domain of \( h^{-1} \) (and the range of \( h \)). This is because applying a function and then its inverse (or vice - versa) undoes the operation of the function.
So, for any \( x \), \( (h\circ h^{-1})(x)=x \).

Step2: Evaluate at \( x = 2 \)

When \( x = 2 \), \( (h\circ h^{-1})(2)=2 \).

Answer:

\( g^{-1}(8)=-4 \)

Part 2: Find \( h^{-1}(x) \)